Let $T$ be a linear transformation on a finite-dimensional vector space $V$ with dimension of $V$ being $d \geq2$. Given that $\text{rank}(T) = 1 $, prove that there exists a scalar $k$ sucht that $T^{2018}$ = $k^{2017}T$.
This was a problem on an exam I had yesterday and there is already a solution posted here, but I'm curious if my reasoning is the same as the solution Kavi Rama provides.
I also said that given that $T$ is unidimensional it sends most vectors in $V$ to zero and that would be the trivial case where $k = 0$. Otherwise, for the only vector that forms a basis of $R(T)$, it is automatically the eigenvector of $T$ and thus set $k$ to be the corresponding $\lambda$.
Should I be using induction instead as the guys over there do? Thank you!
Since the range of $c=k^{2017}$ where $c \in \mathbb{R}$ is $\mathbb{R}$ itself, the question can be reduced to $$T^{2018}x=cx$$ where $c$ is some scalar.
Since $rank(T)=1$, it must knock the whole $V$ into a single line. That is, given any vector $x$, $Tx$ will be located inside the one dimensional column space of $T$. Thus $Tx$ and $T^{2018}x$ must be on the same line. That is, $$T^{2018}x=cx$$ where c is some scalar.
If it is required to present the answer in the form $k^{2017}x$, then we can just extend upon this argument a little. $Tx$ must be an eigenvector of $T$ since $T^2x$ must exist on the same line. If we take the eigenvalue to be $k$, then $T^2x$=$kTx$. It follows that, $$T^{2018}x=k^{2017}Tx$$ and removing x from both sides, $$T^{2018}=k^{2017}T$$ for some scalar $k$